Chapter 16 Playing with Numbers

Explanation:

Here, the addition of first Two digit is 2, 

So taking A as 7 we get, 7+5=12

In which one’s place is 2.

Therefore, A = 7

Now we carry over 1, then

B=3+2+1

B=6 

Hence, A = 7 and B = 6.

Explanation:

Here, the addition of first Two digit is 3, 

So, taking A as 5 we get, 5+8=13

In which one’s place is 3.

Therefore, A = 5

Now we carry over 1, then

CB=4+9+1

CB=14

B = 4 and C = 1

Hence, A = 5, B = 4 and C = 1.

Explanation:

On the place of A we can put 1,2,3,4,5,6,7,8 and so on

But only by multiplying 6×6=36, in which one’s place is 6

Therefore, A = 6 

Explanation:

Starting from the left side we observed that the value of A=3

So, by putting the value of A on the right side we can obtained the value of B=5

Hence, A = 2 and B =5.

Explanation:

Here, on the place of B we can put 0,1,2,3,4,5,6,7,8 and 9

But only the value of B satisfying the equation is 0

Now put the value of A = 5 we get CA=15 

Hence A = 5, B = 0 and C = 1.

Explanation:

Here, on the place of B we can put 0,1,2,3,4,5,6,7,8 and 9

But only the value of B satisfying the equation is 0

Now put the value of A = 5 we get CA=25 

Hence A = 5, B = 0 and C =2

Explanation:

Here on the place of B we can put 4, so 4×6=24

Now carry over 2.

By putting A=7 we can obtain BB=7×6+2=44

Hence, A = 7 and B = 4.

Explanation:

Here we can put B=9 so that 9+1=10

In which one’s place is 0 and carry over is 1.

Now B=9, so by putting A=7 we can obtained the answer as B

Hence, A = 7 and B = 9.

Explanation:

Here putting B = 7, we get 7+1 = 8

Now by putting A = 4, we can obtaine 4+7 = 11

Putting 1 at tens place and carrying over 1, we get

2+4+1 =7

Hence, A = 4 and B = 7.

Explanation:

Here we can assume any value of A from 0-9 and according to that B value can be obtained

So, let’s take A=8, then B=1

So, the addition of left-hand side is 9

And the addition of the middle term is 8+2=10

Putting 0 at one’s place and carry over 1

Now left-hand is 1+6+1=8

Hence, A = 8 and B =1.

Explanation:

Let’s suppose 21y5 is a multiple of 9.

Now according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

That is, 2+1+y+5 = 8+y

Here the value of 8+y can be any multiple of 9 which is 0,9,18,27,36,45 and so on,

But here y is a single digit number so we are taking 8+y=9

So, by solving above equation 

2+1+y+5=9

8+y=9

y=9-8

y=1

Hence the value of y is 1.

Explanation:

Let’s suppose 31z5 is a multiple of 9,

Now according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

That is, 3+1+z+5 = 9+z

Here the value of 9+z can be any multiple of 9 which is 0,9,19,27,36,45 and so on,

This implies, 9+0 = 9 and 9+9 = 18

Hence, 0 and 9 are the two possible answers.

Explanation:

Let’s suppose 24x is a multiple of 3.

Then, according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

That is, 2+4+x = 6+x

Here the value of 6+x is a multiple of 3, and 6+x is one of the numbers: 0, 3, 6, 9, 12, 15, 18 and so on.

Here the value of x should be single digit, which can be 0,3,6 or 9

Thus, x can have any of the four different values: 0 or 3 or 6 or 9.

Explanation:

Let’s suppose 31y5 is a multiple of 3.

Now according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

That is, 3+1+z+5=9+z

Here the value of 9+z can be any multiple of 3 which is 0,3,6,9,12,15 and so on,

At z = 0, 9+z = 9+0 = 9

At z = 3, 9+z = 9+3 = 12

At z = 6, 9+z = 9+6 = 15

At z = 9, 9+z = 9+9 = 18

So, the value of 9+z can be 9,12,15 or 18

Hence 0, 3, 6 or 9 are the four possible answers for z.